Question
The rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ is:
✓
Answer
- $\sqrt{12}+\sqrt{5}$
Solution:
$\frac{1}{2\sqrt{3}-\sqrt{5}}$
$=(2\sqrt{3}-\sqrt{5})(2\sqrt{3}+\sqrt{5})$
$=12-5$
$=7$
Rational number
$(2\sqrt{3}+\sqrt{5})=(\sqrt{4\times3}+\sqrt{5})=\sqrt{12}+\sqrt{5}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The two lines which are parallel to the same line are _______ to each other.
The graph of y + 2 = 0 is a line.
Write the correct answer in the following:
If a linear equation has solutions (-2, 2), (0, 0) and (2, -2), then it is of the form,
If a linear equation has solutions (-2, 2), (0, 0) and (2, -2), then it is of the form,
The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, their volumes are in the ratio:
The triangle is not possible with the given measurements:
The equation of the x-axis is:
-
x = 0
-
y = 0
-
x = y
-
x + y = 0
The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
→- $\frac{1}{2}(\text{ar}\triangle\text{ABC})$
- $\frac{1}{3}(\text{ar}\triangle\text{ABC})$
- $\frac{1}{4}(\text{ar}\triangle\text{ABC})$
- $(\text{ar}\triangle\text{ABC})$
If $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=+1,$ where $\text{x},\ \text{y}\neq0$ then the value of $(x3 - y3)$ is:
→→